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1. Linear Equations
2D : What is a line ?
Informally, a line is a relationship between two
variables (x,y) expressed as y = ax + b,
b, where a
and b are given numbers, eg.
eg. y = 2x + 4,
4, in
which case a=2 and b=4.
b=4.
Pictorially, this looks like this:
Objectives:
• Review and extend familiar ideas
• Special cases (no solution, one solution,
infinitely many solutions)
–Noughty, Chapter 1
–See B&Z (Barnett and Ziegler 4–1)
– See Extended Version of these notes
on the web site
Terminology:
Slope: rate of change (a)
Intercepts:
y-intercept: (0,b)
x-intercept: (-b/a,0)
3
• A system of equations is a set of equations
that we wish to solve simultaneously.
eg
4
1
2¸
y = ax + b
y
9
7
5
3
1
-2
Linear Equations
2
-1
1
2
3
4
5
x
4
Method 1: Graphical
• Noughty Section 1.2
• Example 1.1
– Solve the above system using a graphical
method (like Ex 1, section 4.1, B&Z).
– Hint: The solution is the intersection of the two
lines representing the system.
3 x - 3 y = 3Ô
˝
x + 3y = 7 Ô˛
• A pair of numbers (x0,y0 ) is a solution of
the system if all equations are satisfied by
these values of x and y. Hint: x0=1, y0= 2.
• To solve the system means to find ALL
possible solutions to the system.
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Method 1: Graphical
4
3
x - 13 y = 23 ¸
˝
x + 3y = 7˛
• Here we have one solution.
• Appears to be x = 1, y = 2.
• Check this by substituting in the original
equations.
• When there is only one solution, we say there
is a unique solution.
(A )
y
6
Method 1: Graphical
Solution
y
7/3
Solution
(1,2)
x
-1/2
1/2
7
x
1
Method 2: Algebraic
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We successively produce different systems with
the same solutions, (equivalent systems) until
the solution becomes easy to see.
Noughty Section 1.3
Example 1.1 (again)
The easiest (yet not trivial) 2x2 system is of the
form
x + 0y = a
0x + y = b
x
Given :
=a
y=b
4
3
Wanted :
x - 13 y = 23
x + 3y = 7
x
=a
y=b
Solution (by inspection ) : x=a, y=b.
Method 2: Algebraic
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Noughty, Section 1.3.1
‘Legal’
Legal’
operations
a=?
b =?
Method 2: Algebraic
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Example 1.1 (continued)
We shall use 3 ‘legal’ operations:
4
3
• (A) Interchange two rows.
• (B) Multiply a row by a non-zero constant.
• (C) Add a constant multiple of one row to
another row.
x - 13 y = 23
x + 3y = 7
x - 14 y = 12
x + 3y = 7
x - 14 y = 12
• Convince yourself that these operations are
indeed legal (they do not affect the solution
to the system).
Method 2: Algebraic
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Method 2: Algebraic
13
4
y=
(1)
(2)
(B) Multiply (1) by 3/4
(3)
(4)
(5)
(C) Add -1 times (3) to (4)
13
2 (6)
11
12
Example 1.1 (continued)
x - 14 y =
13
4
y=
1
2
(5)
13
2 (6)
x - 14 y = 12
y=2
x =1
y=2
(7)
(8)
(9)
(10)
(A) Multiply (6) by 4/13
Suggestion
Extr Reading: Section 4-2 B&Z
(don’t worry about reduced matrices)
(C) Add 1/4 of (8) to (7)
x =1
y =2
2
Gauss Jordan Elimination
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Two augmented matrices are called Row
Equivalent if they are augmented matrices of
equivalent systems.
• Noughty Section 1.3.2
• A systematic approach to Method 2, using an
efficient “shorthand” notation.
• We use only the coefficients and take care to
keep them in the correct columns.
• The right hand side (RHS) is included in the
matrix, hence the term Augmented Matrix.
• eg
4
3
x - 13 y = 23
x + 3y = 7
is written as
4
3
We use the symbol
Eg :
Strategy:
Use legal row operations to obtain
Èa
ÍÎ c
b .˘
d .˙˚
È1
ÍÎ0
~
È1 - 14 12 ˘
ÍÎ1 3 7˙˚
~
È1 - 14
ÍÎ0 143
1
2
13
2
˘
˙
˚
Equations (3) & (4)
16
Strategy:
Use legal row operations to obtain
Èa
ÍÎ c
That is, using only the A,B and C operations
make:
“1” in position a
“0” in position c
“1” in position d
“0” in position b
È 43 - 13 23 ˘
ÍÎ1 3 7˙˚
È1 - 14 12 ˘
ÍÎ1
3 7˙˚
Summary
0 .˘
1 .˙˚
Example 1.1(continued)
~
Equations (1) & (2)
15
Summary
~ for this purpose.
È 43 - 13 23 ˘
ÍÎ 1
3 7˙˚
- 13 23
3 7
1
14
Terminology
b .˘
d .˙˚
È1
ÍÎ0
0 .˘
1 .˙˚
Make sure that the new operation
does not mess up what was achieved
by the previous operations.
Example 1.1 (continued)
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Make “1” out of “4/3”
3/4 R1 ‹ R1
(This means: 3/4 x old row
1
gives new row 1)
Make ‘1” in 2nd row into ‘0’.
(–1)R1 +R2 ‹ R2
(old R2 – old R1 gives new R2)
Make ‘13/4”
13/4” in 2nd row into ‘1’.
(4/13)R2 ‹ R2
(4/13 x old R2 gives new R2)
˘
1 3˙
2 ˚
~
È1 - 14
ÍÎ0 143
1
2
~
È1 - 14 12 ˘
ÍÎ0 1 2˙˚
~
È1 0 1 ˘
ÍÎ0 1 2 ˙˚
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Make ‘13/4’
13/4’ in 2nd row into ‘1’.
(4/13)R2 ‹ R2
(4/13 x old R2 gives new R2)
Make ‘-1/4”
-1/4” in 1st row into ‘0’.
R1+(1/4)R2 ‹ R1
(old row 1 + (1/4) old row 2
gives new row 1 )
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Example 1.1 (continued)
~
1x + 0 y = 1
0 x + 1y = 2
È1 0 1 ˘
ÍÎ0 1 2 ˙˚
Always check the results!!!
4
3
19
20
This method is called Gauss Jordan
Elimination.
We say we have reduced the augmented
matrix to one of the form:
x=1,y=2
4
3
x - 13 y = 23
x + 3y = 7
Terminology (See B&Z 4–2)
È1
ÍÎ0
¥ 1- 13 ¥ 2 = 23
1+ 3 ¥ 2 = 7
0 .˘
1 .˙˚
OK
Example 1.2
Example 1.2 (continued)
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2x + y = 1
4x + 2 y = 6
(like Ex 2B,Section 4–
4–1 & Ex 4, section 4–
4–3, B&Z)
2x + y = 1
4x + 2 y = 6
Graphical method (Noughty Page 16):
y
Solve this system of equations.
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No solution, the lines
are parallel.
1
They do not intersect.
3/2
1/2
~
~
È1 12 12 ˘
Í
Î4 2 6 ˙
˚
1
2
1
2
˘
È1
Í
Î0 0 4˙
˚
24
1
2
1
2
˘
È1
Í
Î0 0 4˙
˚
1/2 R1 ‹ R1
–4R1+R2 ‹
R2
x
23
• Gauss Jordan Method :
È2 1 1 ˘
Í
Î4 2 6 ˙
˚
22
Cannot get
È1
ÍÎ0
0 .˘
1 .˙˚
• We can’t get the desired pattern here.
• Whenever you aren’t sure what is going on,
translate back to equations.
• The second row translates to 0x+0y = 4, ie
0 = 4.
• This is impossible fi no solution.
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Example 1.3
25
Example 1.3 (continued)
y
2x + y = 1
4x + 2y = 2
2x + y = 1
4x + 2y = 2
26
1
1/2
x
Graphically:
Solve the system.
The two lines are coincident.
Every point on the line is a solution. Thus
we have an infinitely many solutions –
any (x,y) such that 2x+y=1 is a solution.
Gauss Jordan Method
È2 1 1 ˘
Í
Î4 2 2 ˙
˚
~
~
È1 12 12 ˘
Í
Î4 2 2 ˙
˚
1
2
Gauss Jordan Method
1/2 R1 ‹ R1
–4R1+R2 ‹
R2
1
2
˘
È1
Í
Î0 0 0 ˙
˚
È1 12 12 ˘
Í
Î0 0 0 ˙
˚
• R2 tells us nothing :
0x + 0y = 0, that is, 0 = 0.
• Any (x,y) point such that x+(1/2)y=1/2
is a solution.
Cannot get
È1
ÍÎ0
0 .˘
1 .˙˚
29
È1 12 12 ˘
Í
Î0 0 0 ˙
˚
We shall write the solution in the following way.
Put y = t. Then from the first equation:
x + 12 y =
1
2
x = 12 - 12 y = 12 - 12 t
( x, y ) = ( 12 - 12 t , t )
For every number
t there is a feasible
Solution.
1
2
1
2
30
(x, y) = ( - t, t)
We can generate different solutions by putting
in different values for t.
eg t = 0 gives (x,y) = (1/2,0) as one solution
t = 1 fi
(x,y) = (0,1) as another
solution
t = –3 fi(x,y) = (2,–3)
t = 1/2 fi
(x,y) = (1/4, 1/2) etc
Now do questions 1&2, Sheet 1 in blue booklet.
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Summary
È1
Í
Î0
0 a˘
1 b˙
˚
È1 m n ˘
Í
Î0 0 0 ˙
˚
È1 m n ˘
ÍÎ0 0 p ˙˚
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32
Graphically
We have a unique solution ie
just one solution: (x,y) =(a,b).
We have a unique solution
(ie just one solution).
We have infinitely many
solutions:
We have infinitely many
solutions.
(x,y) = (n–mt,t) , t Œ R
If p ≠ 0, there is no solution
Terminology
There is no solution
Example 1.4
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If a system has solution(s) we say it is
consistent.
If it has no solutions, we say it is inconsistent.
If we have more equations than variables, we
say the system is overdetermined.
If it has fewer equations than variables we say it
is underdetermined.
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Solve the following overdetermined system
graphically.
4x - y = 2
x + 3y = 7
3 x - 4 y = -5
y
7/3
(1,2)
5/4
x
–5/3
1/2
7
–2
Unique solution: (x,y)=(1,2).
Example 1.4 (continued)
4x - y = 2
x + 3y = 7
3 x - 4 y = -5
x=1,y=2
Check this by substitution.
Now do question 3, Example sheet 1.
See 4–3, B&Z
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Larger Systems
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Example 1.5
Solve for x,y,z using Gauss Jordan elimination.
(Like Ex2, Section 4–3 B&Z)
2x + y - 4z = 3
x - 2y + 3z = 4
-3x + 4y - z = -2
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37
Example 1.5
2x + y - 4z = 3
x - 2y + 3z = 4
-3x + 4y - z = -2
È 2
Í 1
ÍÎ-3
1
-2
4
-4 3˘
3 4˙
-1 -2 ˙˚
We want “1” here
È 2
Í 1
ÍÎ-3
1
-2
4
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~
Make “0”
~
Make “1”
-4 3˘ R1 ÷ R2
3 4 ˙ (interchange
-1 -2 ˙˚ rows 1 and 2 )
~
Make “0”
È 1 -2 3 4 ˘
Í 2 1 -4 3 ˙ –2R1+R2 ‹ R2
Í
Î-3 4 -1 -2 ˙
˚ 3R1+R3 ‹ R3
È1 -2 3 4 ˘
Í0 5 -10 -5˙ 1/5 R2 ‹ R2
Í
Î0 -2 8 10 ˙
˚
È1 -2 3 4 ˘ 2R2+R1 ‹ R1
Í0 1 -2 -1˙
Í
Î0 -2 8 10˙
˚ 2R2+R3 ‹
R3
Gauss Jordan
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È1 0 -1 2 ˘
Í0 1 -2 -1˙
Make “1” Í
Î0 0 4 8 ˙
˚ 1/4 R3 ‹ R3
The systematic way to do this is to work from
left to right, along the diagonal, creating the
desired pattern (or as near to it as we can get):
~
È1 0 -1 2 ˘ R3+R1 ‹ R1
Í0 1 -2 -1˙ 2R3+R2 ‹ R2
Make “0” Í0 0 1 2 ˙
Î
˚
a b
d e
g h
~
~
~
È1 0 0 4˘ Thus:
Í0 1 0 3˙ x=4, y=3, z=2
Í
Î0 0 1 2˙
˚
Check it!
Partial Adjustment
Noughty Section 1.3.3 . Sometime it is better to
leave the upper diagonal entries as they are:
Make “0”
È1 -2 3 4 ˘
Í0 1 -2 -1˙ 2R2+R3 ‹
R3
Í0 -2 8 10˙
Î
˚
È1 -2 3 4 ˘
Í0 1 -2 -1˙
ÍÎ0 0 4 8 ˙˚ 1/4 R3 ‹
R3
Make “1”
~
~
È1 -2 3 4 ˘
Í0 1 -2 -1˙
ÍÎ0 0 1 2 ˙˚
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1 0 0
0 1 0
0 0 1
c
f
i
Be careful at every stage not to destroy the
pattern already created.
Partial Adjustment
41
42
We now solve the system by successive
substitution (bottom-up):
Eg:
Eg:
È1 -2 3 4 ˘
Í0 1 -2 -1˙
ÍÎ0 0 1 2 ˙˚
z=2
y-2z =-1
x-2y+3z=4
x - 2 y + 3z = 4
y - 2 z = -1
z=2
y-2(2) =-1
x-2(3)+3(2)=4
(x,y,z) = (4,3,2)
y=3
x=4
Check it!
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Example 1.6
(see exercise 3, section 4–
4–3 B&Z)
The following augmented matrix arises at
some stage in solving a system of equations:
È1 2 3 4 ˘ What does this tell us about
Í
˙ the solution?
0 0 0 10
ÍÎ1 -1 -1 1 ˙˚ R2 gives 0x + 0y + 0z = 10
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Note
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Do not use the Speeding Up procedure described in
Noughty, Section 1.3.4.
The description is there just FYI.
This is not possible,
possible, so the system has
no solution.
solution. There is no need to do any
further reduction.
(Now do all of Q4 & 5, on Example sheet 1)
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